$-\dfrac{6}{4} + \dfrac{3}{8} = {?}$
Solution: $ = - {\dfrac{6 \times 2}{4 \times 2}} + {\dfrac{3 \times 1}{8 \times 1}} $ $ = - {\dfrac{12}{8}} + {\dfrac{3}{8}} $ $ = - \dfrac{{12} + {3}}{8} $ $ = -\dfrac{9}{8}$